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I'm studying some topics about topological games of length $\alpha\geq\omega$, where I came across the following statement about additively indecomposable ordinals (recall that $\alpha$ is additively indecomposable if $\beta+\gamma <\alpha$ whenever $\beta,\gamma<\alpha$):

If $\alpha\geq\omega$ is additively indecomposable, then we may write $\alpha=\bigcup_{i<cof(\alpha)} Y_i$, where $Y_i$ and $Y_j$ are disjoint if $i<j$ , $Y_i$ has order type $\alpha$ for each $i$, and $\{min(Y_i):i<cof(\alpha)\}$ is cofinal in $\alpha$.

Although it seems to me intuitively correct (at least for countable ordinals), I could not formulate an argument to prove that such a construction is always possible. So, my question is: how can I prove the above statement?

Thank you!

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For the countable case, this is very easy: let $\alpha$ be countable (so $cf(\alpha)=\omega$) and additively decomposable. Then $\alpha$ is a limit, and we can decompose $\alpha$ into its "odd" and "even" bits: that is, there is a function $f: \alpha\rightarrow\alpha$ which is injective and order-preserving, such that $f(0)>0$, and $\alpha - im(f)\cong\alpha$. By iterating $f$, we get the desired partition of $\alpha$: $Y_0=\alpha- im(f)$, $Y_{n+1}=\alpha - (f^{n+1}(\alpha)\cup \bigcup_{i\le n} Y_i).$ (The condition $f(0)>0$ is used to guarantee that the minima of the $Y_i$ are cofinal in $\alpha$.)


For uncountable $\alpha$, I find it easier to switch to a slightly less concrete picture. Let $Y_i=\alpha\times\{i\}$ for $0\le i<cf(\alpha)$, and let $(\gamma_i)_{i<cf(\alpha)}$ be an increasing cofinal subset of $\alpha$ with $\gamma_0=0$. Let $h: cf(\alpha)^2\cong cf(\alpha)$ be a pairing function on the cofinality of $\alpha$ such that $$(*): \,\, \text{ $\{h(0, i): i<cf(\alpha)\}$ is cofinal in $\alpha$,}$$ and for $i=h^{-1}(j, k)$ let $$L_i=[\gamma_j, \gamma_{j+1}]\times\{k\}.$$ for $i<cf(\alpha)$. Then by induction, $\sum_{i<j} L_i<\alpha$ for each $j<\alpha$: at limit stages we use the cofinality of $\alpha$, and at successor stages we use the additive indecomposability of $\alpha$. On the other hand, we clearly have $\alpha\le\sum_{i<cf(\alpha)}L_i$, and so $$ \alpha\cong\sum_{i<cf(\alpha)}L_i.$$ This representation on $\alpha$ then pulls back to a collection of maps from the $T_i$, which - by our condition $(*)$ on $h$ - clearly satisfies that the images of the minima are cofinal.

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  • $\begingroup$ Basically, the general construction just carves $\alpha$ into $cf(\alpha)$-many "blocks," and dovetails which block comes from which $T_i$. $\endgroup$ – Noah Schweber Oct 16 '13 at 21:15
  • $\begingroup$ Noah: When I accepted your answer I focused only in the countable case (since it was the one I needed); by the way, thank you for that. Today, looking again to the general (uncountable) case with more attention, some questions arise to me: Is the pairing function $h$ related to Gödel's pairing function? And, at the end of your proof, what are the $T_i$ sets? Thanks again. $\endgroup$ – Renan Maneli Mezabarba Feb 15 '14 at 15:57
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    $\begingroup$ $h$ is just any bijection $cf(\alpha)^2\cong cf(\alpha)$ such that $\{h(0, i): i< cf(\alpha)\}$ is cofinal in $\alpha$ for all $\alpha$; there doesn't have to be any relation to Goedel's pairing function. As to the $T_i$, that's a typo - "$T_i$" should be "$Y_i$". $\endgroup$ – Noah Schweber Feb 15 '14 at 18:53

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